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Chapter 6: Problem 486

\(\forall n \in N,\left(3+5^{(1 / 2)}\right)^{n}+\left(3-5^{(1 / 2)}\right)^{\mathrm{n}}\) is (a) Even natural number (b) Odd natural number (c) Any natural number (d) Rational number

Short Answer

Expert verified
The expression \(\left(3+5^{(1/2)}\right)^n+\left(3-5^{(1/2)}\right)^n\) simplifies to \(\sum_{m=0}^{\lfloor n/2 \rfloor} 2\binom{n}{2m} 3^{n-2m} 5^{m}\), which is always an even natural number. Therefore, the correct answer is (a) Even natural number.

Step by step solution

01

Expanding using the binomial theorem

: Applying the binomial theorem to the given expression, we get: \((3+5^{(1/2)})^n+(3-5^{(1/2)})^n = \sum_{k=0}^n \binom{n}{k} 3^{n-k} (5^{(1/2)})^k + \sum_{k=0}^n \binom{n}{k} 3^{n-k} (-1)^k (5^{(1/2)})^k\)
02

Combining the sums

: We can simplify this by combining the sums: \[= \sum_{k=0}^n \binom{n}{k} 3^{n-k} (5^{(1/2)})^k + \binom{n}{k} 3^{n-k} (-1)^k (5^{(1/2)})^k\] \[= \sum_{k=0}^n \binom{n}{k} 3^{n-k} (5^{(1/2)})^k (1+(-1)^k)\] Notice when \(k\) is even, the value of \((-1)^k\) is 1, and when \(k\) is odd, its value is -1. Thus, the \((1+(-1)^k)\) term will be 0 for odd \(k\) and 2 for even \(k\). Therefore, only even \(k\) values will contribute to the sum.
03

Simplifying the sum

: As only even \(k\) contribute, we can rewrite the sum using \(k=2m\), where \(m\) is an integer: = \(\sum_{m=0}^{\lfloor n/2 \rfloor} 2\binom{n}{2m} 3^{n-2m} (5^{(1/2)})^{2m}\) Now we can simplify the expression: = \(\sum_{m=0}^{\lfloor n/2 \rfloor} 2\binom{n}{2m} 3^{n-2m} 5^{m}\)
04

Analyzing the expression

: Now let's analyze the simplified expression for evenness, oddness, and rationality. - Evenness: The expression will always be even, as each term in the sum contains a factor of 2. So, the expression is an even natural number. - Rationality: All terms in the sum are integers since each term is a product of an integer binomial coefficient and positive integer powers of 3 and 5. Therefore, the resulting sum is also an integer, and thus a rational number. So, the correct answer is (a) Even natural number.

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